Calculating number of bits in a cache

Question

Preface: There are many different design patterns that are important to cache's overall performance. Below are listed parameters for different direct-mapped cache designs.

• Cache data size: 32 kib
• Cache block Size: 2 words
• Cache access time: 1-cycle

Question: Calculate the number of bits required for the cache listed above, assuming a 32-bit address. Given that total size, find the total size of the closest direct-mapped cache with 16-word blocks of equal size or greater. Explain why the second cache, despite its larger data size, might provide slower performance that the first cache.

Here's the formula:
Number of bits in a cache 2^n X (block size + tag size + valid field size)

Here's what I got: 65536(1+14X(32X2)..
is this correct?

Solution 1

using: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))

for the first one i get:

total bits = 2^15 (1+14+(32*2^1)) = 2588672 bits

for the cache with 16 word blocks i get:

total bits = 2^13(1 +13+(32*2^4)) = 4308992

the next smallest cache with 16 word blocks and a 32 bit address works out to be 2158592 bits, smaller than the first cache.

Solution 2

I'm stuck on the same problem too but I have the answer to the first part.

To calculate the total number of bits required

1. You need to convert the KB to words and get the index bits.
2. Use the answer from part 1 to get your tag bits.

3. Plug them into this formula.

   (2^(index bits)) * ((tag bits)+(valid bits)+(data size))
   

Hint: data size is 64 bits in this case and valid bit is 1. So just find the index and tag bits.

And I don't think your answer is right. I didn't check but I can see you are multiplying 1+14 and (32x2) instead of adding them.

Comments

[Chittolina]

How did you find 2^15 blocks on the first part? Since the cache data size is 32KiB and there is 2 words/block, 32 * 2^10 bytes / 8 bytes results in 2^12 blocks.

[Peter Cordes]

The question doesn't specify the word size, but given that address-size is 32-bit, a good guess would be word-size = 32-bit. (So a pointer is 1 word, like a typical 32-bit RISC). More likely than the 16-bit word size you're assuming. A better question would have specified the word size instead of leaving answerers to guess. Perhaps it was implied by some context the OP left out.